# Find the determinant of a particular matrix without a calculator.

I would like to know the trick to solving this determinant problem.

Find the determinant of the matrix $$\begin{bmatrix} 283&5&\pi&347.86\times10^{15^{83}}\\ 3136 & 56 & 5 & \cos(2.7402)\\ 6776 & 121 & 11 & 5\\ 2464 & 44 & 4 & 2 \end{bmatrix}$$ Hint: do not use a calculator.

This is a problem here (section 3.2). According to it, the answer is 6.

I have no idea how to do this. Obviously some of the entries in this matrix are red herrings, and I need to perform some trick. But the only manipulations I know towards computing the determinant is adding one row to another and multiplying a row by a scalar, and nothing in this matrix suggests that I do that.

I've tried decomposing the matrix as follows, $$\begin{bmatrix} 283 & 5 & \pi & 347.86\times10^{15^{83}}\\ 56^2 & 56 & 5 & \cos(2.7402)\\ 11^2\cdot 56 & 11^2 & 11 & 5\\ 4\cdot11\cdot 56 & 4\cdot 11 & 4 & 2 \end{bmatrix},$$

though I'm not sure what I accomplished (I could also note that $$56 = 14\cdot 4, 4 = 2^2$$ etc. and do more decomposing). The problem is not all entries in any column/row are nice, and the nicest looking ones are the first and second column (or third and fourth row). Maybe the matrix is the product of two nice ones?

• You already know column one is about $56$-times column two, so perform a column reduction operation to reduce the entries in column one substantially. Similarly, row four is almost $4/11$-times row three, so perform a row reduction. Can you find enough reductions (without dragging $\pi$ and the other inconvenient numbers out of their rows/columns) to make the matrix upper diagonal (or at least have so many zeroes in the left two columns and bottom two rows that the alternating sum of weighted minors formula for the determinant has almost all terms equal to zero)? Commented Aug 24, 2021 at 1:43

HINT: Note that $$\begin{bmatrix}283 \\ 3136 \\ 6776 \\ 2464\end{bmatrix} - 56 \begin{bmatrix}5 \\ 56 \\ 121 \\ 44\end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ which allows you to replace the leftmost column by $$[3, 0, 0, 0]^T$$. You should be able to then write the determinant of this $$4\times4$$ matrix in terms of the determinant of a smaller matrix.
• Wow... it does, this leaves the matrix $\begin{bmatrix}56&5&\cos(2.7402)\\121&11&5\\44&4&2\end{bmatrix}$, then note $\begin{bmatrix}56\\121\\44\end{bmatrix}-11\begin{bmatrix}5\\11\\4\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$...! Commented Aug 24, 2021 at 1:58